# How can I show that the polynomial $p = x^5 - x^3 - 2x^2 - 2x - 1$ is irreducible over $\Bbb Q$?

I've tried a few "criteria" to check if this is irreducible. According to Maple it only has one entirely real root which I suspect is not rational but I can't prove it so I'm attempting to check if $p$ is irreducible.

Eisenstein's Criterion doesn't work here and I'm yet to find a suitable transformation such that it could work. I also read that if a polynomial is irreducible over $\Bbb F_q$, with $q$ a prime not dividing the leading coefficient, then it is irreducible over $\Bbb Q$ so I reduced the polynomial modulo $2$ to obtain

$$p \equiv x^5 + x^3 + 1 \mod 2.$$

I think this is correct but then I need to know how to check the irreducibility of this new polynomial over $\Bbb F_2$. Do I simply need to check that neither $0$ nor $1$ are roots of this polynomial? (And am I applying this theorem correctly?)

If this polynomial IS irreducible over $\Bbb Q$, is the splitting field obtained by simply adjoining the roots to $\Bbb Q$?

• Since the polynomial is of degree $5$ it might be possible for it to have no roots in $\mathbb{Q}$, but be reducible. Writing it as a product of polynomials of degree $2$ and $3$ might be possible. Apr 9, 2017 at 11:11
• To answer your last question: yes. This follows straight from the definition of a splitting field.
– user583416
Apr 9, 2017 at 11:18
• Oh, I've found something wonderful. Answer coming up. Apr 9, 2017 at 11:19

.A little bit of scouting for nice irreducibility criteria throws up some very nice results:

Here is a lovely lemma by (Prof.) Ram Murty:

Let $$f(x) = a_mx^m + ... + a_1x + a_0$$ be a polynomial of degree $$m$$ in $$\mathbb Z[x]$$. Let $$H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$$. If $$f(n)$$ is prime for some $$n \geq H+2$$, then $$f(x)$$ is irreducible in $$\mathbb Z[x]$$.

In our case, $$a_m = 1$$, and the maximum of all the quantities in question is $$2$$. Hence, if $$f(n)$$ is prime for some $$n \geq 4$$, then we are done.

You can check that for $$n=4$$, the number $$f(4) =919$$, which is prime!

Hence, it follows that the polynomial is irreducible.

ASIDE : There is also a "shifted" base (base shifts from $$0...n-1$$ to $$|b| < \frac n2$$) version of Cohn's criteria, which will tell you that if $$f(10)$$ is prime, then the given polynomial is irreducible. This matches that description, since all coefficients are between $$-5$$ and $$5$$. Very interestingly, $$f(10) = 98779$$ is also prime! (Hence, another proof by another wonderful result).

• That is very cool! Thanks for the link. Apr 9, 2017 at 11:28
• You are welcome. I've seen some of these criteria very long ago, while doing a Galois theory polynomial "scouting mission". Apr 9, 2017 at 11:31
• Nice method. Just wondering, why not going already with $f(4)=919$ which is also a prime?
– Sil
Apr 26, 2018 at 23:30
• You are right, it works out. I was doing that calculation mentally, so I skipped directly to $n=5$ because I must have made some mistake at $f(4)$. Apr 26, 2018 at 23:34

By Gauss' Lemma we have that the polynomial is irreducible over $\mathbb{Q}$ if and only if it's irreducible over $\mathbb{Z}$. Now the easiest way would be to prove that polynomial is irreducible over $\mathbb{Z}_2$, which would be enough.

Assume it's reducible. As the polynomial has no roots over $\mathbb{Z}_2$, then the only possibility is if it's a product of polynomials of degree $2$ and $3$. So assume that:

$$x^5 + x^3 + 1 = (x^3 + ax^2 + bx + c)(x^2 + dx + e) \quad \quad \text{over } \quad\mathbb{Z}_2$$

Then multiply everything out and compare the factors. As $a,b,c,d,e \in \mathbb{Z}_2$, you only have two options. Eventually you will get that $c=e=1$ and $a=d=b$. But this would imply that $c+bd + ea = a^2 + a + 1 = 1$ in $\mathbb{Z}_2$. But this is impossible, as it's the coefficient in front of $x^2$ and it should be $0$.

Hence the polynomial is irreducible over $\mathbb{Z}_2$ and eventually $\mathbb{Z}$ and $\mathbb{Q}$

• I think I'll accept this one as being the most useful in a general setting, thanks a lot. Apr 9, 2017 at 11:28
• One could also just divide by the only irreducible polynomial in $\mathbb{Z_2}[x]$ which is $x^2+x+1$ and see it cannot be the factor.
– Sil
Aug 11, 2018 at 9:23
• @Sil True. However I doubt that an OP asking factorizing of a polynomial is aware that $x^2+x+1$ is the only irreducible quadratic in $\mathbb{Z}_2$ Aug 11, 2018 at 9:25

Your proof is almost complete: it is indeed sufficient to prove that $p(x)\in \mathbb F_2[x]$ (the reduction mod $2$ of your polynomial) is irreducible.
The polynomial $p(x)$ has no factor of degree $1$ since it has no zero in $\mathbb F_2$, so there remains only to prove that $p(x)$ has no factor which is an irreducible polynomial $g(x)\in \mathbb F_2[x]$ of degree $2$.
But the only such irreducible polynomial is $g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved.

By the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\pm1$. Clearly, these are not roots.

Similary, any quadratic factor would have to be $x^2+ax\pm1$ with $a\in\Bbb Z$. You might be able to exclude these manually ...